The Quantum Rotor

It is useful to simply investigate angular momentum with just one free rotation angle. This might be the quantum plane propeller. We will do a good job of this in 3 dimensions later.

Lets assume we have a mass $m$ constrained to move in a circle of radius $r$. Assume the motion in the circle is free, so there is no potential. The kinetic energy is \bgroup\color{black}${1\over 2}mv^2={p^2\over 2m}$\egroup for \bgroup\color{black}$p=mr{d\phi\over dt}$\egroup.

If we measure distance around the circle, then \bgroup\color{black}$x=r\phi$\egroup and the one problem we have is that once I go completely around the circle, I am back to \bgroup\color{black}$x=0$\egroup. Lets just go ahead and write our wavefunction.

\begin{displaymath}\bgroup\color{black} e^{i(px-Et)/\hbar}= e^{i(pr\phi-Et)/\hbar} \egroup\end{displaymath}

Remembering angular momentum, lets call the combination \bgroup\color{black}$pr=L$\egroup. Our wave is \bgroup\color{black}$e^{i(L\phi-Et)/\hbar}$\egroup.

This must be single valued so we need to require that

e^{i(2\pi L-Et)/\hbar}=e^{i(0-Et)/\hbar} \\
e^{i(2\pi L)/\hbar}=1 \\
L=n\hbar \\
n=0,1,2,3... \\

The angular momentum must be quantized in units of $\hbar$.

This will prove to be true for 3 dimensions too, however, the 3 components of angular momentum do not commute with each other, leading to all kinds of fun.

Jim Branson 2013-04-22